University Physics IV · Nuclear Physics · 13.1
Nuclear Size & the Charge Form Factor
Electrons are the cleanest ruler we have for a nucleus: no strong force, a vertex QED gives exactly, and a resolution you set with the beam energy. This is how a diffraction pattern in a spectrometer becomes a radius, a surface thickness, and a density that refuses to move.
Build the model
Connect the measurement to the mechanism.
A nucleus cannot be looked at; it can only be bounced off, and the pattern read. Electrons are the right projectile because they feel only the electromagnetic current, whose vertex QED gives exactly, so anything unexpected in the data belongs to the target and not to the probe. Drive them hard enough that the resolution ħ/q falls below a fermi and the scattering stops looking point-like: the cross-section departs from the Mott prediction, and the ratio of the two defines the form factor F(q).
In first Born approximation F is the Fourier transform of the charge density — one dimensionless function of momentum transfer, normalised to F(0) = 1, carrying shape and no Z at all. Its slope at small q returns ⟨r²⟩ with no model assumed; its minima return a radius; a wide enough sweep in q inverts to ρ(r) itself. What comes back is a two-parameter Fermi profile whose half-density radius grows as A¹⁄³ while its interior sits near 0.16 nucleons per fm³ for everything from carbon to uranium.
That saturation is the first hard evidence that the nuclear force is short-ranged and self-limiting, and the liquid-drop model is built on it. The cost is written into the method: Born maps charge, so this locates protons rather than nucleons, and at Z = 82 the expansion parameter Zα = 0.60 is not small, so the single Fourier transform must give way to a full phase-shift analysis.
- Simple definition
- The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1.
- Example
- For ²⁰⁸Pb at 250 MeV the first minimum of |F|² sits near θ = 29°, where q/ħ = 2E sin(θ/2)/ħc = 0.634 fm⁻¹; setting qR/ħ = 4.493 gives R = 7.08 fm, within 0.4% of 1.2 × 208¹⁄³ = 7.11 fm.
The probe resolves detail of order ħ/q, so 0.634 fm⁻¹ sees 1.6 fm — a fermi needs hundreds of MeV.
E in MeV for E ≫ mₑc², recoil neglected; q/ħ in fm⁻¹. 250 MeV at 29° gives 0.634 fm⁻¹.
This is the denominator you divide the data by. It is smooth in θ, so every feature that survives belongs to the nucleus.
αħc = 1.440 MeV fm; cos²(θ/2) is electron spin, the bracket is target recoil, Mc² the nuclear rest energy.
One curve replaces every beam energy: 250 and 500 MeV data must fall on the same F plotted against q.
ρch in e fm⁻³; F is dimensionless with F(0) = 1, so Z cancels and F carries shape only.
The one model-independent number in the subject: ¹²C returns √⟨r²⟩ = 2.47 fm from that slope alone.
The expansion holds while qR/ħ ≲ 1; the slope of F against q² as q → 0 assumes no profile at all.
One minimum angle fixes R with no curve fitting; the spacing of later minima then tests the shape you assumed.
R is the equivalent uniform radius, R = √(5/3) √⟨r²⟩. Measured minima are filled in, never true zeros.
Only c moves with A. A fixed interior density is saturation, and it is the assumption the liquid-drop model rests on.
c ≈ 1.1A¹⁄³ fm, a ≈ 0.55 fm, so t ≈ 2.4 fm; ρ₀ ≈ 0.16 nucleons fm⁻³ and A cancels.
Why an electron, and how hard you have to hit
An electron has no strong interaction and no measurable size of its own, so the only thing coupling it to a nucleus is the electromagnetic current, whose vertex QED gives exactly. Anything unexpected in the data therefore belongs to the target. What the beam energy buys is resolution: the probe sees structure of order ħ/q, and since q/ħ = 2E sin(θ/2)/ħc for E ≫ mₑc², with ħc = 197.3 MeV fm, a momentum transfer of 0.634 fm⁻¹ resolves only 1.6 fm. Getting below a fermi means hundreds of MeV — Hofstadter's Stanford measurements ran at 200 to 550 MeV, and it is no accident that nuclear structure appeared in scattering data only once accelerators reached that range. One further number must be checked before any of this is used: the Born expansion parameter Zα. For ¹²C it is 6/137 = 0.044 and single-photon exchange is an excellent approximation. For ²⁰⁸Pb it is 82/137 = 0.60, and it is not.
Divide out Mott, and keep what refuses to cancel
The point-charge prediction is Rutherford with two relativistic repairs. Electron spin supplies a factor cos²(θ/2), because helicity conservation forbids exact backscattering of an ultrarelativistic electron, and target recoil supplies 1/[1 + (2E/Mc²) sin²(θ/2)]. Both are smooth and monotonic in θ: the Mott cross-section has no minima, no shoulders, nothing an experimenter could mistake for structure. Nor is the recoil factor doing heavy lifting here — at 250 MeV and 29° it amounts to 1.6 × 10⁻⁴ for ²⁰⁸Pb and 2.8 × 10⁻³ for ¹²C. So divide the measured elastic cross-section by the Mott value point by point in angle. The ratio is |F(q)|²: it starts at 1 in the forward direction and falls three or four orders of magnitude by 30°, with minima on the way down. That fall is the entire measurement.
Read the transform three different ways
For a spherically symmetric density the angular integral in F(q) collapses the exponential to sin(qr/ħ)/(qr/ħ), and three readings follow. Expand for small argument: F ≈ 1 − q²⟨r²⟩/6ħ², so the initial slope of F against q² gives the mean-square charge radius with no profile assumed — the only genuinely model-independent number here. Next, the positions of the minima fix a radius, once a shape is assumed. Third, data spanning a wide enough range of q can be inverted directly to ρ(r), which is what produced the charge-density curves in every textbook figure. A consistency test comes free with the first Born approximation: F depends on q alone, so 250 MeV data and 500 MeV data must land on one curve when plotted against q rather than against θ. Where they separate, Born is failing and the extracted radius is not trustworthy.
Spacing gives the radius, envelope gives the surface
For a uniform sphere F(x) = 3(sin x − x cos x)/x³ with x = qR/ħ, vanishing at x = 4.493, 7.725 and 10.90 — the zeros of a circular aperture, and for the same diffraction reason. One minimum angle therefore fixes R, and the spacing of the later minima tests whether a sphere was a fair description. It is not quite. Real data show the maxima between minima dying away much faster than a sphere predicts, because it is a sharp edge that keeps a diffraction pattern alive. Softening that edge over a diffuseness a multiplies the envelope by roughly exp(−πqa/ħ), which at a = 0.55 fm and q/ħ = 2 fm⁻¹ is a factor of 0.03. Spacing measures the radius; the rate of decay measures the surface. Two independent numbers out of one pattern is exactly why a two-parameter profile is enough.
Saturation: 0.138 and 0.160 are both A-independent
Fits return the two-parameter Fermi form ρ(r) = ρ₀/[1 + exp((r − c)/a)] with a ≈ 0.55 fm for essentially everything, so the 90%-to-10% surface thickness t = 4a ln3 = 2.4 fm hardly moves from ¹⁶O to ²³⁸U while c grows as A¹⁄³. Two densities fall out of that, and they are not the same. Normalising, A = (4π/3)ρ₀c³[1 + π²(a/c)²], which for ²⁰⁸Pb with c = 6.62 fm and a = 0.55 fm gives ρ₀ = 0.160 nucleons fm⁻³ — the interior value, with light nuclei within about 10% of it. Separately, the equivalent uniform sphere R = √(5/3)√⟨r²⟩ follows R ≈ 1.2A¹⁄³ fm, a fit good to a few percent above A ≈ 90 that drifts to 1.39A¹⁄³ at carbon; its mean density is A/[(4π/3)(1.2)³A] = 0.138 nucleons fm⁻³, with A cancelling identically. Both constants say one thing — nucleons pack at a fixed density because the force saturates — and they differ by 15% only because the equivalent radius is pushed outward to cover a diffuse tail.
What the Born approximation cannot hand you
Three limits are built into the method. First, F is the transform of the charge density, so electron scattering locates protons; the neutron distribution needs another probe, and the parity-violating PREX measurement puts the ²⁰⁸Pb neutron rms radius about 0.28 fm outside the proton one. Second, what is measured is ρch, which already has the proton's own 0.84 fm charge radius folded into it; that has to be unfolded before the result is a point-nucleon distribution. Third, one-photon exchange assumes Zα ≪ 1, and at Z = 82 it is 0.60: the incoming and outgoing waves are distorted by the nuclear Coulomb field, and the measured minima are shifted inward by several percent and partly filled in. A defensible heavy-nucleus analysis is an iterated phase-shift calculation against a trial ρ(r), not a single Fourier transform. Quoting a lead radius from a plane-wave fit is a systematic error, not a rounding one.
Change one variable at a time
Make the relationship visible.
Hold a = 0.55 fm, set A = 208, c = 6.60 fm: the plateau lands on 0.162 fm⁻³. Pull c to 5.50 fm and it leaps to 0.272, denser than any nucleus. Now try A = 16, c = 2.60 fm, a = 0.50 fm: c is 60% smaller yet the plateau holds at 0.159. Saturation — c tracks A¹⁄³, the interior does not.
INTERIOR DENSITY0.162 fm⁻³
c / A¹⁄³1.114 fm
SURFACE t = 4a ln32.42 fm
UNIFORM R = 1.2A¹⁄³7.11 fm
Live interpretationINTERIOR DENSITY: 0.162 fm⁻³. c / A¹⁄³: 1.114 fm. SURFACE t = 4a ln3: 2.42 fm. UNIFORM R = 1.2A¹⁄³: 7.11 fm
Catch the common trap
Explain before calculating.
At one fixed beam energy, elastic electron scattering puts the first diffraction minimum at a larger angle for ¹⁶O than for ²⁰⁸Pb. What does that tell you?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUse R = 1.2A¹⁄³ fm to find the radius of ²⁰⁸Pb and of ¹⁶O, then the mean nucleon density of each, and convert one of them to kg m⁻³.
- 208¹⁄³ = 5.925, so R(²⁰⁸Pb) = 1.2 × 5.925 = 7.11 fm. 16¹⁄³ = 2.520, so R(¹⁶O) = 1.2 × 2.520 = 3.02 fm.
- Volumes from V = (4/3)πR³: for lead, 4.1888 × 7.11³ = 4.1888 × 359.4 = 1.506 × 10³ fm³; for oxygen, 4.1888 × 27.54 = 115.4 fm³.
- Mean nucleon densities: 208/1506 = 0.138 fm⁻³ and 16/115.4 = 0.139 fm⁻³. They agree because R³ ∝ A makes A cancel identically, leaving 1/[(4π/3)(1.2)³].
- In SI units: 0.138 nucleons fm⁻³ × 1.6605 × 10⁻²⁷ kg × 10⁴⁵ fm³ m⁻³ = 2.3 × 10¹⁷ kg m⁻³, some 2 × 10¹⁴ times the density of water.
AnswerR(²⁰⁸Pb) = 7.11 fm and R(¹⁶O) = 3.02 fm; both give 0.138 nucleons fm⁻³, or 2.3 × 10¹⁷ kg m⁻³. Nuclear matter does not compress as A grows.
MediumA 250 MeV electron beam scatters elastically from ¹²C. At θ = 10.0° the measured cross-section is 0.905 of the Mott value. Find the rms charge radius from that single point, and say which way the truncated expansion pushes the answer.
- Momentum transfer: q/ħ = 2E sin(θ/2)/ħc = 2(250)(sin 5.00°)/197.33 = 500 × 0.087156/197.33 = 0.2208 fm⁻¹, so q²/ħ² = 0.04877 fm⁻².
- The measured ratio is |F|², so F = √0.905 = 0.9513. F is real and positive here, well short of the first zero.
- Small-q expansion: F ≈ 1 − q²⟨r²⟩/6ħ², so ⟨r²⟩ = 6(1 − F)ħ²/q² = 6 × 0.04868/0.04877 = 5.989 fm².
- Take the square root: √⟨r²⟩ = √5.989 = 2.45 fm. As a check on the method, R = √(5/3) × 2.45 = 3.16 fm, so qR/ħ = 0.2208 × 3.16 = 0.70 — comfortably inside the range where the expansion is usable.
- The next term is +q⁴⟨r⁴⟩/120ħ⁴ and it is positive, so the true F exceeds the linear form and the truncation returns a radius that is too small. At qR/ħ ≈ 0.70 that bias is about 1%, and the accepted ¹²C value is 2.47 fm.
Answer√⟨r²⟩ = 2.45 fm from one point, about 1% under the accepted 2.47 fm — the bias a quadratic truncation always carries. Remove it by fitting the slope of F against q² as q → 0.
HardElastic scattering of 250 MeV electrons from ²⁰⁸Pb puts the first minimum of |F|² at θ = 29.0°. Treat the nucleus as a uniform sphere to extract R, check it against 1.2A¹⁄³, then compare that sphere's mean density with the interior density of the fitted Fermi profile c = 6.62 fm, a = 0.55 fm.
- q/ħ = 2(250) sin(14.5°)/197.33 = 500 × 0.25038/197.33 = 0.6344 fm⁻¹.
- The first zero of 3(sin x − x cos x)/x³ is at x = qR/ħ = 4.493, so R = 4.493/0.6344 = 7.08 fm. Against 1.2 × 208¹⁄³ = 7.11 fm that is 0.4%, and √(3/5) × 7.08 = 5.49 fm for the rms, against the accepted 5.50 fm.
- Mean density of that sphere: V = (4/3)π(7.08)³ = 4.1888 × 354.9 = 1488 fm³, so 208/1488 = 0.140 nucleons fm⁻³.
- Normalise the Fermi profile instead: A = (4π/3)ρ₀c³[1 + π²(a/c)²] = 4.1888 × 290.1 × [1 + 9.870 × (0.55/6.62)²] ρ₀ = 1215 × 1.0681 ρ₀ = 1298 ρ₀.
- So ρ₀ = 208/1298 = 0.160 nucleons fm⁻³, 15% above the equivalent sphere's 0.140, because that sphere's radius is stretched outward to reproduce ⟨r²⟩ across a 2.4 fm surface.
AnswerR = 7.08 fm, within 0.4% of 1.2A¹⁄³ = 7.11 fm. The equivalent sphere averages 0.140 nucleons fm⁻³ while the real interior is ρ₀ = 0.160 fm⁻³. Both are A-independent; they are not the same number.